Question

When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.

Answer 1

$\lambda =400nm=400\times {10}^{-9}m$

$\because \frac{hc}{\lambda }=\frac{hc}{{\lambda }_0}+e{v}_0$

$\Rightarrow \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{{\lambda }_0}+1.6\times {10}^{-19}\times 1.1$

$\Rightarrow \frac{6.63\times 3}{4}\times \frac{{10}^{-34}\times {10}^8}{{10}^{-7}}$ $=\frac{6.63\times 3\times {10}^{-26}}{{\lambda }_0}+1.6\times 1.1\times 10-19$

$\Rightarrow 4.97\times {10}^{-19}$

$=\frac{19.89\times {10}^{-26}}{{\lambda }_0}+1.76\times {10}^{-19}$

$\Rightarrow 4.97=\frac{19.89\times {10}^{-7}}{{\lambda }_0}+1.76$

$\Rightarrow \frac{19.89\times {10}^{-7}}{{\lambda }_0}=4.97-1.76=3.21$

$\Rightarrow {\lambda }_0=\frac{19.89\times {10}^{-7}}{3.21}$

$=6.196\times {10}^{-7}m=620nm$